C passing array of pointers - c

Question

This relates to C. I am having some trouble understanding how I can assign strings to char pointers within arrays from a function.
#include <stdio.h>
#include <string.h>
void function(char* array[]);
int main(void)
{
char* array[50];
function(array);
printf("array string 0: %s\n",array[0]);
printf("array string 1: %s\n",array[1]);
}
void function(char* array[])
{
char temp[] = "hello";
array[0] = temp;
array[1] = temp;
return;
}
Ideally, I would like the main printf function to return
array string 0: hello
array string 1: hello
But I'm having trouble understanding arrays of pointers, how these pass to functions and how to manipulate them in the function. If I declare a string like char temp[] = "string" then how do I assign this to one of the main function array[i] pointers? (assuming I have my jargon right)

Answer 1

char temp[] = "hello"; only creates a local, temporary array inside the function. So when the function exists, the array will be destroyed.
But with array[0] = temp; you're making array[0] point to the local array temp.
After the function returns, temp doesn't exist anymore. So accessing array[0] which pointed to temp will cause undefined behavior.
You could simply make temp static, so it also exists outside the function:
static char temp[] = "hello";
Or, you could copy the "hello" string to array[0] and array[1]. For copying C-strings, you normally use strcpy.
char temp[] = "hello";
strcpy(array[0], temp);
strcpy(array[1], temp);
However, before copying you need to make sure array[0] and array[1] point to memory that has enough space to hold all characters of "hello", including the terminating null character. So you have to do something like this before calling strcpy:
array[0] = malloc(6);
array[1] = malloc(6);
(6 is the minimum numbers of characters that can hold "hello".)

Answer 2

how do I assign this to one of the main function array[i] pointers
Arrays cannot be assigned.
A pointer cannot hold an array, it can only refer to an array. For the latter the pointer needs to get an array's address assigned.
Referring 1.
This
char temp[] = "hello";
isn't an assigment, but an initialisation.
This
char temp[];
temp[] = "hello";
would not compile (the 2nd line errors), as an array cannot be assigned.
Referring 2.
This
char* array[50];
defines an array of 50 pointers to char, it could reference 50 char-arrays, that is 50 C-"strings". It cannot hold the C-"strings" themselfs.
Example
Applying what is mentioned above to your code whould lead to for example the following:
#include <stdio.h>
#include <string.h>
void function(char* array[]);
int main(void)
{
char* array[50];
/* Make the 1st two elements point to valid memory. */
array[0] = malloc(42); /* Real code shall test the outcome of malloc()
as it might fail and very well return NULL! */
array[1] = malloc(42);
function(array);
printf("array string 0: %s\n",array[0]);
printf("array string 1: %s\n",array[1]);
return 0;
}
void function(char* array[])
{
char temp[] = "hello";
/* Copy the content of temp into the memory that was allocated to
the array's 1st memebrs in main(). */
strcpy(array[0], temp);
strcpy(array[1], temp);
return;
}

Answer 3

first, you need to allocate the destination.
second, char temp[] = "hello"; in function() is local variable. you cannot use their outside of the function. use strcpy to copy the content.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void function(char* dest)
{
char temp[] = "hello";
strcpy(dest, temp);
return;
}
int main(void)
{
// or just char dest[10] = {0};
char *dest = malloc(10);
function(dest);
printf("dest: %s\n", dest);
}
In you program, you defined char* array[50];, so you need to create memory space for each item:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void function(char* a[])
{
char temp[] = "hello";
strcpy(a[0], temp);
strcpy(a[1], temp);
return;
}
int main(void)
{
char *a[50];
int i = 0;
for (i = 0; i < 50; ++i)
a[i] = malloc(10);
function(a);
printf("a[0]: %s\n", a[0]);
printf("a[1]: %s\n", a[1]);
}